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ladessa [460]
3 years ago
9

How many grams are in 9.30 x 10-2 moles of calcium phosphate, Caz(PO3)2?

Chemistry
1 answer:
irina [24]3 years ago
6 0

Answer:

25.9 g

Explanation:

Given that,

No of moles of calcium phosphate, n=9.3\times 10^{-2}

We need to find how many grams of calcium phosphate has this much of no of moles.

Mass divided by molar mass is equal to the no of moles on a molecule.

The molar mass of calcium phosphate is 310.18 g/mol

Using the concept of no of moles as follows :

n=\dfrac{m}{M}\\\\m=n\times M\\\\m=9.3\times 10^{-2}\times 310.18\\\\m=28.84\ g

Out of given options, option (a) i.e. 25.9 g is the correct answer.

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How many moles are in a 111.5 gram sample of sodium chloride, NaCl?
Katarina [22]

Answer: 1.91 moles

Explanation:

First you want to find the molar mass of NaCL

Na = 22.99g  Cl = 35.45g

22.99g + 35.45g = 54.44g

Then divide 111.5g by 54.44g and this will give you moles.

5 0
3 years ago
What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso
Paraphin [41]

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

 

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M 
[A-]= 0.609/ 2.00 = 0.3045 M 
pKa = 6.25 

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

6 0
3 years ago
What pressure is exerted by 0.883 mol N, in a 4.68 L steel container at 197.9 °C?
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6 0
3 years ago
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g )
Kisachek [45]

Answer:

After complete reaction, 0.280 moles of ammonia are produced

Explanation:

Step 1: Data given

Number of moles N2 = 0.140 moles

Number of moles H2 = 0.434 moles

Step 2: The balanced equation

N2(g) + 3H2 (g) ⟶ 2NH3 (g)

Step 3: Calculate the limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. It will completely be consumed (0.140 moles).

H2 is in excess. There will react 3*0.140 = 0.420 moles

There will remain 0.434 - 0.420 = 0.014 moles

Step 4: Calculate moles NH3

For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3

After complete reaction, 0.280 moles of ammonia are produced

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3 years ago
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